The number of critical points of the function | vedclass

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Question

(B) For $x 
eq 0$,$f(x) = |\frac{\sin x}{x}|$. Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. $1$. At $x = 0$,$f(x)$ is continuous

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) For $x 
eq 0$,$f(x) = |\frac{\sin x}{x}|$. Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. $1$. At $x = 0$,$f(x)$ is continuous and has a local maximum,so $x = 0$ is a critical point. $2$. For $x 
eq 0$,$f(x) = 0$ when $\sin x = 0$,which gives $x = \pm \pi$ in the interval $(-2\pi, 2\pi)$. At these points,the function is not differentiable because of the absolute value,so $x = \pm \pi$ are critical points. $3$. We also check for $f'(x) = 0$ where $\frac{\sin x}{x} 
eq 0$. This requires $\frac{x \cos x - \sin x}{x^2} = 0$,or $	an x = x$. The roots of $	an x = x$ (other than $x=0$) are approximately $\pm 4.49$,which lie outside the interval $(-2\pi, 2\pi)$ since $2\pi \approx 6.28$. Thus,the critical points in $(-2\pi, 2\pi)$ are $x = 0, \pi, -\pi$. Total number of critical points is $3$.

The number of critical points of the function $f(x) = \begin{cases} |\frac{\sin x}{x}|, & x \ne 0 \\ 1, & x = 0 \end{cases}$ in the interval $(-2\pi, 2\pi)$ is equal to:

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